# Necessary and sufficient conditions for differentiability.

Apologizes if I'm missing something in my question or if my question seems trivial; this is my first question on this site. As motivation for my question, consider the following standard first year calculus question.

Consider this piecewise function: $f(x) = \left\{ \begin{array}{lr} ax^2+b & \text{ if } x \le-2\\ 12x-5 & \text{ if } x >-2 \end{array} \right.$

For what values of $a$ and $b$ will $f(x)$ be differentiable?

To solve this question, I would like to propose the following theorem:

$\mathbf{Theorem:}$ A function $f(x)$ is differentiable iff $f'(x)$ is continuous.

If this theorem is true, then I can solve for $a$ first by noting that: $f'(x) = \left\{ \begin{array}{lr} 2ax & \text{ if } x \le-2\\ 12 & \text{ if } x >-2 \end{array} \right.$

Thus, since by my theorem $f'(x)$ must be continuous, we have:

\begin{align*} \lim_{x \rightarrow -2^-}f'(x) &= \lim_{x \rightarrow -2^+}f'(x)\\ \lim_{x \rightarrow -2^-}2ax &= \lim_{x \rightarrow -2^+}12\\ 2a(-2) &= 12\\ -4a &= 12 \\ a &= -3 \\ \end{align*}

Hence, since differentiability implies continuity, we can solve for $b$ as follows:

\begin{align*} \lim_{x \rightarrow -2^-}f(x) &= \lim_{x \rightarrow -2^+}f(x)\\ \lim_{x \rightarrow -2^-}-3x^2+b &= \lim_{x \rightarrow -2^+}12x-5\\ -3(-2)^2+b &= 12(-2)-5\\ b-12 &= -29\\ b &= -17 \\ \end{align*}

so that our differentiable function is:

$$f(x) = \left\{ \begin{array}{lr} -3x^2-17 & \text{ if } x \le-2\\ 12x-5 & \text{ if } x >-2 \end{array} \right.$$

Anyways. My question is: Is my proposed theorem actually a thing? I've looked through my calculus textbook and it doesn't seem to explicitly state it, yet I don't know how to solve this question otherwise. If this theorem turns out to be false, how else can you solve this problem? Thanks in advance. =]

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A differentiable function need not have a continuous derivative. – user10444 May 9 '13 at 20:50
$f(x)$ is a number, not a function. So your statement should be "$f$ is differentiable if and only if $f'$ is continuous". It should be clear what the problem is here. When you speak of the derivative $f'$, you are already assuming that $f$ is differentiable. – wj32 May 9 '13 at 20:53
On the other hand, one can show: if $\lim_{x\rightarrow c^-} f'(x)$ exists, then the derivative from the left of $f$ at $x=c$ exists and is equal to the value of this limit. A similar fact holds for the derivative from the right. Also, a function is differentiable at $x=c$ if and only if both one-sided derivatives exist at $c$ and are equal there. So your method for computing $a$ is correct, but not for the reason you give. – David Mitra May 9 '13 at 21:05
@DavidMitra Thanks, that makes perfect sense. – Adriano May 9 '13 at 23:15

let $f(x)= \begin{cases} x^2\sin(\frac 1 x), & \mbox{if } x \not= 0 \\0 & \mbox{if } x=0 \end{cases}$ is continuous but has a discontinuous derivative. Check the continuity of $f'(x)$ at $x=0$.

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Sorry, but how does this contradict my (admittedly false) theorem? I don't think your function $f(x)$ is differentiable at $x = 0$ (that is, I don't think that $f'(0)$ exists). This agrees with my (admittedly false) theorem since, as you said, $f'(x)$ is discontinuous at $x = 0$. – Adriano May 9 '13 at 22:28
$f'(0)$ exists because $$\lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2\sin(1/h)}{h} = \lim_{h \to 0} h\sin(1/h) = 0.$$ – zrbecker May 9 '13 at 22:37
Ah, yes, of course. Then we have: $$f'(x) = \left\{ \begin{array}{lr} 2x \sin(1/x)-\cos(1/x) & \text{ if } x \ne 0\\ 0 & \text{ if } x = 0 \end{array} \right.$$ While $f'(0)$ exists, $f'$ is still not continuous since \lim_{x \rightarrow 0}f'(x) doesn't exist. – Adriano May 9 '13 at 22:46

It is not. A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable iff

$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$

exists for all $x\in\mathbb{R}$. If $f'$ is continuous, $f$ is said to be continuously differentiable (or of class $\mathcal{C}^1$). Hence, to solve the problem you need to pick $a$ and $b$ such that the limit exists for any $x\in\mathbb{R}$. Note also, that for $f$ to be differentiable, $f$ must be continuous.

By the way, welcome to math stackexchange and I wouldn't worry about asking a "trivial question" - your question is well written and, often, questions are only trivial if you've already seen the answer.

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For that function to be differentiable you need $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ to exist.

The only point where something "weird" is going on, is at -2. Thus we just need to find $a, b$ such that $$\lim_{h \to 0^+} \frac{f(-2 + h) - f(-2)}{h} = \lim_{h \to 0^-} \frac{f(-2 + h) - f(-2)}{h}$$

The right hand side is simply $-4a$, by using the regular rules of derivatives. The weirdness happens on the other side of the limit.

$$\lim_{h \to 0^-} \frac{12(-2 + h) - 5- (4a + b)}{h} = \lim_{h \to 0^-} \frac{-29 - 4a -b + 12h}{h}$$

In order for this limit to exist we need $-29 -4a -b = 0$. Otherwise the limit will be $-\infty$ or $+\infty$. Once we have $-29 -4a -b = 0$, we get $$\lim_{h \to 0^-} \frac{-29 - 4a -b + 12h}{h} = \lim_{h \to 0^-} \frac{12h}{h} = 12.$$

Since we need the left and the right limit to be the same, we get $-4a = 12$. Thus $a = -3$. Then solving for $b$, we have $-29 + 12 - b = 0$, so $b = -17$.

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I think you meant to say that $$\lim_{h \to 0^+} \frac{12(-2 + h) - 5- (4a + b)}{h} = \lim_{h \to 0^+} \frac{-29 - 4a -b + 12h}{h}= \lim_{h \to 0^+} \frac{12h}{h} =12$$ so that $-4a=12$ gives $a=-3$ and $-29-4(-3)-b=0$ yields $b=-17$. While your method works, I'm not really convinced why you were able to deduce that $-29-4a-b=0$ though. – Adriano May 9 '13 at 22:09
The way I went about it was to say, if $-29 - 4a - b = c$ where $c \neq 0$, then $c/h \to \pm \infty$. So the limit doesn't exist. But the limit must exist for it to be differentiable. – zrbecker May 9 '13 at 22:22
Another way to think about it, is $f$ needs to be continuous at $-2$, so $a(-2)^2 + b = 12(-2) - 5$, which after some algebra gives $-29 - 4a - b = 0$. – zrbecker May 9 '13 at 22:25
Ah that makes sense now, thanks. – Adriano May 9 '13 at 22:29